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Energy required to thaw one gallon of ice

by eniwetok
Tags: energy, gallon, required, thaw
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eniwetok
#1
Dec15-10, 12:21 PM
P: 3
Not sure if the title does this question justice. My question how much energy in KWHs or a fraction thereof, could be saved in refrigerator operation if someone took a gallon of water, let it freeze solid outside during a 0F degree night, then placed it in their refrigerator until it thawed to around 35F roughly the ambient temperature in the fridge.

Thanks!
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NobodySpecial
#2
Dec15-10, 12:47 PM
P: 474
The energy needed to heat a material is Q = mcT
m is the mass, T is the temperature change and c is the specific heat capacity (in metric units water=4.2KJ/kg/deg C ice=2KJ/kg/deg C)

Then the energy needed to melt the ice (and so the energy given out when ice melts) is called the latent heat of fusion ( 334 kJ/kg for ice->water)
eniwetok
#3
Dec15-10, 01:02 PM
P: 3
Quote Quote by NobodySpecial View Post
The energy needed to heat a material is Q = mcT
m is the mass, T is the temperature change and c is the specific heat capacity (in metric units water=4.2KJ/kg/deg C ice=2KJ/kg/deg C)

Then the energy needed to melt the ice (and so the energy given out when ice melts) is called the latent heat of fusion ( 334 kJ/kg for ice->water)
Thanks for responding NS, but you're talking to someone who managed to get through college but hadn't passed a math class since junior HS algebra. LOL

NobodySpecial
#4
Dec15-10, 01:44 PM
P: 474
Energy required to thaw one gallon of ice

Ok working in metric - cos I don't know the numbers for lbs/fahenheit etc

To heat 1 kg of ice from -18C (approx 0f) to 0C gives Q = 1Kg * 2KJ/kg/degC * 18degC = 36KJ
To melt 1 kg of ice gives Q = 1Kg * 334 KJ/kg = 334KJ
Then heating the water form 0C to 2C (ie 35F) gives Q = 1Kg * 4.2KJ/kg/degC * 2C = 8.4KJ

So overall you get 36KJ + 334KJ + 8.4KJ = 378KJ
378/3600 = 0.1 KWh
eniwetok
#5
Dec15-10, 02:01 PM
P: 3
Quote Quote by NobodySpecial View Post
So overall you get 36KJ + 334KJ + 8.4KJ = 378KJ
378/3600 = 0.1 KWh
Thanks NS... now that makes sense. So it's the same amount of electricity as running a 100 watt bulb for an hour. And at a 10 cents a KWH Guess I was hoping for more savings than a penny. Oh well. Thanks again!
DaleSwanson
#6
Dec15-10, 03:01 PM
P: 351
Quote Quote by eniwetok View Post
Thanks NS... now that makes sense. So it's the same amount of electricity as running a 100 watt bulb for an hour. And at a 10 cents a KWH Guess I was hoping for more savings than a penny. Oh well. Thanks again!
While it is true that it will take 0.1 kWh to warm the ice, that does not mean that it will only consume 0.1 kWh from the wall. That would be the case only if the refrigerator was 100%, which it is not. Unfortunately, I don't have even a ballpark figure for how efficient refrigerators are. While a google search brings up lots of hits, they are mainly concerned with more practical measures of efficiency, and I'm unsure how to convert it.

Anecdotally, my experience with coolers is that if you were to replace a gallon's worth of ice on a daily basis they should be able to keep cool indefinitely. While refrigerators are larger, it suggests to me that it could, at least, help.

It would be rather easy to test with an experiment. Measure the refrigerator's energy consumption for a week, then add a gallon of ice every day for a week and measure it again.
russ_watters
#7
Dec15-10, 07:39 PM
Mentor
P: 22,301
Quote Quote by DaleSwanson View Post
While it is true that it will take 0.1 kWh to warm the ice, that does not mean that it will only consume 0.1 kWh from the wall. That would be the case only if the refrigerator was 100%, which it is not. Unfortunately, I don't have even a ballpark figure for how efficient refrigerators are.
Actually, since refrigerators are moving energy around, not creating useful work, they are not rated in terms of efficiency, but rather COP and they typically have a COP of 2.5:1. So the reality is the refrigerator will use a lot less than .1 kWh to make that ice.


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